QUESTION 6
Easy
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
SOLUTION
1
Form the AP
First term(initial salary), \(a = 5,00,000\)
Common difference(increment), \(d = 20,000\)
So the AP is: \(5,00,000, \,5,20,000,\, 5,40,000,...\)
Common difference(increment), \(d = 20,000\)
So the AP is: \(5,00,000, \,5,20,000,\, 5,40,000,...\)
2
Use the \(n^{th}\) term formula
Let the \(n^{th}\) term, \(a_n = 7,00,000\)
\(a_n = a + (n − 1)d\)
⇒ \(7,00,000 = 5,00,000 + (n − 1)(20,000)\)
⇒ \(2,00,000 = (n − 1)(20,000)\)
⇒ \(10 = n − 1\)
⇒ \(n = 11\)
Thus, ₹7,00,000 is the \(11^{th}\) salary amount.
Since the first salary corresponds to year 1, his income reaches ₹7,00,000 after 10 years.
\(a_n = a + (n − 1)d\)
⇒ \(7,00,000 = 5,00,000 + (n − 1)(20,000)\)
⇒ \(2,00,000 = (n − 1)(20,000)\)
⇒ \(10 = n − 1\)
⇒ \(n = 11\)
Thus, ₹7,00,000 is the \(11^{th}\) salary amount.
Since the first salary corresponds to year 1, his income reaches ₹7,00,000 after 10 years.
🏆
Final Answer : His income reaches ₹7,00,000 after 10 years.
Concept Note
The \(n^{th}\) term of an AP is given by:
\(a_n = a + (n − 1)d\)
where,
\(a = \)first term
\(d = \)common difference
\(n = \) term number
To find when the salary reaches ₹7,00,000, we treat ₹7,00,000 as the \(n^{th}\) term and substitute the known values into the formula, and solve for \(n\).